How Do You Calculate Recoil Shift in Atomic Emissions?

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SUMMARY

This discussion focuses on calculating the recoil shift for atomic emissions, specifically the most energetic Lyman line from a free hydrogen atom and the emission of a 14.4 keV gamma from a free Fe-57 atom. The recoil shift is calculated using the formula ΔE = E² / 2mc², where E represents energy, m is mass, and c is the speed of light. The user successfully calculated the recoil shift for the Fe-57 gamma emission as 0.002 eV but expressed uncertainty regarding the Lyman line calculation, indicating a potential error in their initial energy calculation.

PREREQUISITES
  • Understanding of atomic physics and emission spectra
  • Familiarity with the Lyman series and its energy calculations
  • Knowledge of the recoil shift formula ΔE = E² / 2mc²
  • Basic proficiency in unit conversions, particularly between keV and MeV
NEXT STEPS
  • Review the calculation of the most energetic Lyman line using E = hc / λ
  • Study the implications of recoil shifts in atomic emissions
  • Explore the differences in energy scales between keV and MeV emissions
  • Learn about the properties of Fe-57 and its gamma emissions
USEFUL FOR

Students and professionals in physics, particularly those studying atomic emissions, recoil effects, and energy calculations in quantum mechanics.

atomicpedals
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Homework Statement



b) Calculate the recoil shift for the most energetic lyman line from a free H atom.

c) Calculate the recoil shift for the emission of a 14.4keV gamma from a free atom of Fe-57.

Homework Equations



E = hc / \lambda = h \nu

\Delta E = E^2 / 2mc^2

The Attempt at a Solution



I'm still not sure I'm going about this correctly...

b) start by finding the most energetic lyman line E = hc / \lambda = (1.981E-16)/(912A) = 2.17E-19 (this is where I think my error is). Then find the recoil shift
\Delta E = E^2 / 2mc^2 = (4.72E-38)/(0.003) = 1.57E-35.

c) calculate the recoil shift directly using the mass for Fe-57
\Delta E = E^2 / 2mc^2 = (207.36)/(1.89E-22)(2.99E8)^2 = 12.27MeV ...again this answer just feels wrong, how can a keV scale emission result in an MeV scale recoil?

Quite clearly I've made some horrible mistake in here, I need some help to find what I did wrong.
 
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Ok, I've been able to answer part c:

((14.4keV)^2)/2(53.02GeV) = 0.002eV

(sorry for not using tex with the equation, I've been spending way too much time with C++ lately)

Still working on part b though.
 

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